Problem: Solve for $x$, $ -\dfrac{2x - 4}{2x - 4} = \dfrac{8}{2x - 4} - \dfrac{2}{8x - 16} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2x - 4$ $2x - 4$ and $8x - 16$ The common denominator is $8x - 16$ To get $8x - 16$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{2x - 4}{2x - 4} \times \dfrac{4}{4} = -\dfrac{8x - 16}{8x - 16} $ To get $8x - 16$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{2x - 4} \times \dfrac{4}{4} = \dfrac{32}{8x - 16} $ The denominator of the third term is already $8x - 16$ , so we don't need to change it. This give us: $ -\dfrac{8x - 16}{8x - 16} = \dfrac{32}{8x - 16} - \dfrac{2}{8x - 16} $ If we multiply both sides of the equation by $8x - 16$ , we get: $ -8x + 16 = 32 - 2$ $ -8x + 16 = 30$ $ -8x = 14 $ $ x = -\dfrac{7}{4}$